Math Kangaroo Solutions 2012 Level 7-8 and Math Kangaroo 2018 Level 7-8

Q24 - David wants to arrange the twelve numbers from 1 to 12 in a circle so that any two neighboring numbers differ by either 2 or 3. Which of the following pairs of numbers have to be neighbors?

A)  5 and 8            B)  3 and 5            C)  7 and 9          D)  6 and 8            E)  4 and 6

S24 - First, we know that 1 and 12 must be directly across from each other. Next, we find that the numbers next to 1 are 3 and 4 while the numbers next to 12 are 9 and 10. Then, we place the remaining numbers on there and see that, with trial and error, 3 and 9 must be on one side and 4 and 10 must be on the other. Then, we can place the remaining numbers to get a working circle:

1
3        4
6                  2
8                        5
11                 7
9        10
12

We check all the combos and only D) 6 and 8 works.

Q26 - A book contains 30 stories each starting on a new page. The lengths of the stories are 1, 2, 3, ..., 30 pages. The first story starts on the first page. What is the largest number of stories that can start on an odd-numbered page?

A)  15           B)  18            C)  20            D)  21            E)  23

S26 - We see that the first page is odd-numbered. Next, we add an even number to get another odd-numbered page. But if we add an odd number, we get an even numbered page. So we get the following sequences:

- OOOOOOOOOOOOOOOEOEOEOEOEOEOEOE = 22 Odds
- OEOEOEOEOEOEOEOOOOOOOOOOOOOOOO = 23 Odds

Then, we get E) 23 since it is the most.

Q27 - A rope is folded in half, then in half again, and then in half again. Finally the folded rope is cut through, forming several strands. The lengths of two of the strands are 4 m and 9m. Which of the following could not have been the length of the whole rope?

A)  52 m           B)  68 m            C)  72 m            D)  88 m            E)  All the previous are possible.

S27 - We must try all of the choices, finding the folded length by dividing by 8.

----|---------
----|---------       These are just 3 folds and they are connected on the ends ( actually in a V shape )
----|---------

For 52, we get the folded length is 6.5 and can be split into 2 and 4.5, so at the folds, they are 4 which is 2*2 and 9 which is 4.5*2.

For 68, we get the folded length is 8.5 which can be split into 4 and 4.5, where 4 is one end of the rope and 4.5 is part of a fold to get 4.5*2 = 9

For 72, we get the folded length is 9, but we can't figure out any way to split it since we can't do 4.5 nor 2 on any fold with the correct length on then ends.

Just to try the last one, 88, we get the folded length is 11 and we can do 2 at the fold and 9 at the end to get 2*2 = 4

So our answer is C) 72 m.

Q30 - A positive number need to be placed in each cell of the 3*3 grid shown, so that in each row and each column the product of the three numbers is equal to 1, and in each 2*2 square the product of the four numbers is equal to 2. What number should be placed in the central cell?

A)  16            B)  8            C)  4            D)  1/4            E)  1/8

S30 - We can assign the letters a-i to the grid, and use the two properties to find which squares must contain the same numbers and solve for e.

abc
de  f
ghi

Since each 2*2 square has a product of 2, we can see that (a*b*d*e)*(d*e*g*h) = 2*2 = 4. We also know that each row and each column has a product of 1, which means that (a*d*g)*(b*e*h) = 1*1 = 1, so (a*b*d*e)*(d*e*g*h) = (a*d*g)*(b*e*h) * (d*e) = 1*1*(d*e) = 4.

We can apply this logic to all the middle side squares d, b, f, h, which means that (d*e) = (b*e) = (f*e) = (h*e) = 4. So, since the row of (d*e*f) = 1, we have that (d*e)*(f*e) = (d*e*f)*e = 1*e = e = 16.

So, our answer is the central cell must have A) 16.

Q: Domino tiles are said to be arranged correctly if the number of dots at the ends that touch are the same. Paulius laid six dominoes in a line as shown in the diagram. He can make a move by either swapping the position of any two dominoes or by rotating one domino. What is the smallest umber of moves he needs to make to arrange all the tiles correctly?

(A) 1     (B) 2    (C) 3     (D) 4     (E) It is impossible to do


Solution:
___   ___    ___   ___    ___   ___
|4|6|   |3|1|   |4|2|   |3|4|   |6|1|   |2|6|

First, we can swap the third and fifth dominoes to get:
___   ___    ___   ___    ___   ___
|4|6|   |3|1|   |6|1|   |3|4|   |4|2|   |2|6|

Next, we can swap the second and third dominoes to get:
___   ___    ___   ___    ___   ___
|4|6|   |6|1|   |3|1|   |3|4|   |4|2|   |2|6|

Rotating the third domino, we get
___   ___    ___   ___    ___   ___
|4|6|   |6|1|   |1|3|   |3|4|   |4|2|   |2|6|

Which is a valid configuration, so the answer is (C) 3


Q: Points N, M, and L lie on the sides of the equilateral triangle ABC, such that NM is perpendicular to BC, ML is perpendicular to AB, and LN is perpendicular to AC. The area of triangle ABC is 36. What is the area of triangle LMN?

(A) 9     (B) 12     (C) 15     (D) 16     (E) 18

Solution:

We can see that each of the triangles bordering LMN are 30-60-90 triangles because of the 60 degree angle of the equilateral triangle and a 90 degree angle because of the perpendicular lines.
We can then see that the LMN is an equilateral triangle and label one of the sides x. Then, we get the side length of the ABC is 3x/√3, using this fraction, we can square it to find the ratios of the areas is 1/3 to get the area of LMN is (B) 12.

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